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Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

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alok kumar singh

Contributor-Level 10

$\begin{array}{l} (i) {| a z_{1} ? b z_{2} |}^{2} + {| b z_{1} + a z_{2} |}^{2} \\ ? ? ? ? ? ? = {| a z_{1} |}^{2} + {| b z_{2} |}^{2} ? 2 R e (a z_{1} . b {\bar{z}}_{2}) + {| b z_{1} |}^{2} + {| a z_{2} |}^{2} + 2 R e (a z_{1} . b {\bar{z}}_{2}) \\ ? ? ? ? ? ? = {| a z_{1} |}^{2} + {| b z_{2} |}^{2} + {| b z_{1} |}^{2} + {| a z_{2} |}^{2} \\ ? ? ? ? ? ? = (a^{2} + b^{2}) ({| z_{1} |}^{2} + {| z_{2} |}^{2}) \\ ? ? ? ? ? ? ? ? H e n c e, ? ? t h e ? ? v a l u e ? ? o f ? ? t h e ? ? f i l l e r ? ? i s ? ? (a^{2} + b^{2}) ({| z_{1} |}^{2} + {| z_{2} |}^{2}) . \\ (i i) \sqrt[]{? 25} * \sqrt[]{? 9} = \sqrt[]{? 1} . \sqrt[]{25} * \sqrt[]{? 1} \sqrt[]{9} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 5 i * 3 i = 15 i^{2} = ? 15 \\ ? ? ? ? ? ? ? H e n c e, ? ? t h e ? ? v a l u e ? ? o f ? ? t h e ? ? f i l l e r ? ? i s ? ? ? 15 . \end{array}$

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Maths Matrices

Let A be a matrix of order 2 * 2, whose entries are form the set {0, 1, 2, 3, 4, 5}. If the sum of all the entries of A is a prime number p, 2 < p < 8, then the number of such matrices A is.

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Payal Gupta

Contributor-Level 10

Sum of all entries of matrix A must be prime p such that 2 < p < 8 then sum of entries may be 3, 5 or 7

If sum is 3 then possible entries are

(0, 5), (0, 1, 4), (0, 2, 3), (0, 1, 3)

(0, 1, 2, 2) and (1, 2)

$∴$ Total number of matrices = 4 + 12 + 12 + 12 + 12 + 4 = 56

If sum is 7 then possible entries are

(0, 2, 25), (0, 03, 4), (0, 1, 5), (0, 3, 1), (0, 2, 3), (1, 4), (1, 2, 2), (1, 2, 3) and (0, 1, 2, 4)

Total number of matrices with sum 7 = 104

$∴$ total number of required matrices

= 20 + 56 = 104

= 108

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a year ago

Maths Complex Numbers and Quadratic Equations

Let $α, β$ , be the roots of the equation $x^{2} - 4 λ x + 5 = 0 a n d α, γ$ be the roots of the equation $x^{2} - (3 \sqrt[]{2} + 2 \sqrt[]{3}) x + 7 + 3 λ \sqrt[]{3} = 0,$ If $β + γ = 3 \sqrt[]{2},$ then ${(α + 2 β + γ)}^{2}$ is equal to.

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Payal Gupta

Contributor-Level 10

$α, β$ are roots of x² - $4 λ x + 5 = 0$

$∴ α + β = 4 λ a n d α β = 5$

Also $α, γ$ are roots of

$x^{2} - (3 \sqrt[]{2} + 2 \sqrt[]{3}) x + 7 + 3 \sqrt[]{3} λ = 0, λ > 0$

$∴ α + γ = 3 \sqrt[]{2} + 2 \sqrt[]{3}, α γ = 7 + 3 \sqrt[]{2 λ}$

$? α$ is common root

$∴ α^{2} - 4 λ α + 5 = 0$ …….(i)

And

$α^{2} - (3 \sqrt[]{2} + 2 \sqrt[]{3}) α + 7 + 3 \sqrt[]{3} λ = 0$ ….(ii)

From (i) – (ii) ; we get

$α = \frac{2 + 3 \sqrt[]{3} λ}{3 \sqrt[]{2} + 2 \sqrt[]{3} - 4 λ}$

$? β + γ = 3 \sqrt[]{2}$

$∴ {(α + 2 β + γ)}^{2} = {(α + β + β + γ)}^{2}$

$= 98$

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a year ago

Maths Relations and Functions

Let S = ${1, 2, 3, 4, 5, 6, 7, 8, 9, 10} .$ Define f : S S as

$f (n) = {2 n & , i f n = 1, 2, 3, 4,$
$2 n - 1$
$\begin{matrix} i f n = 6, 7, 8, 9, 10 \end{matrix} .$

Let g:S S be a function such that log(n) = ${\begin{matrix} n + 1 & , \\ i f n i s o d d & n - 1 \\ , & i f n i s e v e n \end{matrix},$ Then $g (10) (g (1) + g (2) + g (3) + g (4) + g (5))$ is equal to.

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Payal Gupta

Contributor-Level 10

. $? f (n) = {\begin{matrix} 2 n, n = 1, 2, 3, 4, 5 \end{matrix}$

$\begin{matrix} 2 n - 11, n = 6, 7, 8, 9, 10 \end{matrix}$

$∴ f (1) = 2, f (2) = 4, . . . ., f (5) = 10$

And f(6) = 1, f(7) = 3, f(8) = 5, …., f(10) = 9

Now,

$f (g (n)) = {\begin{matrix} n + 1 & i f n i s o d d \\ n - 1, & i f n i s e v e n \end{matrix}$

$∴ f (g (10)) = 9 \Rightarrow g (1) = 1$

$∴ g (10) (g (1) + g (2) + g (3) + g (4) + g (5)) = 190$

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a year ago

Maths Statistics

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and $\frac{9}{4}$ respectively. Then $x^{4} + y^{2}$ is equal to.

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Payal Gupta

Contributor-Level 10

$= \frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$

$∴ x + y = 12$ …. (i)

And variance

$= \frac{2^{2} + 1^{2} + 0^{2} + 0^{2} + 1^{2} + 2^{2} + {(x - 6)}^{2} + {(y - 6)}^{2}}{8}$

$= \frac{9}{4}$

$∴ {(x - 6)}^{2} + {(y - 6)}^{2} = 8$ ……. (ii)

From (i) and (ii)

x = 4 and y = 8

$∴ x^{4} + y^{2} = 320$

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a year ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Which of the following statement is a tautology?

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Payal Gupta

Contributor-Level 10

$? ((\sim q) \land p) \lor (p v (\sim p))$

$= (\sim q \land p) \lor t (t i s)$

$\equiv t$

$∴$ option (C) is correct.

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a year ago

Maths Trigonometric Functions

a = sin 36° is a root of which of the following equation?

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Payal Gupta

Contributor-Level 10

$α = s i n 36 ° = x (s a y)$

$∴ x = \frac{\sqrt[]{10 - 2 \sqrt[]{5}}}{4}$

$\Rightarrow 16 x^{2} = 10 - 2 \sqrt[]{5}$

$\Rightarrow 16 x^{4} - 80 x^{2} + 20 = 0$

$∴ 4 x^{4} - 20 x^{2} + 5 = 0$

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a year ago

Maths Maths Integrals

The value of $c o t (\sum_{n = 1}^{50} t a n^{- 1} (\frac{1}{1 + n + n^{2}})) i s$

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Payal Gupta

Contributor-Level 10

$c o t (\sum_{n = 1}^{50} t a n^{- 1} (\frac{1}{1 + n + n^{2}}))$

$= c o t (t a n^{- 1} 51 - t a n^{- 1} 1)$

$= c o t (c o t^{- 1} (\frac{52}{50}))$

$= \frac{26}{25}$

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a year ago

Maths Probability

If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x – 6y = 30, then the probability that y < 1 is

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Payal Gupta

Contributor-Level 10

The required probability

$= \frac{A r e a o f R e g i o n P Q C A P}{A r e a o f R e g i o n A B C A}$

$= \frac{\frac{1}{2} * 8 * 6 - \frac{1}{2} * 2 * 4}{\frac{1}{2} * 8 * 6}$

$= \frac{5}{6}$

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Maths Vector Algebra

Let $\vec{a} a n d \vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt[]{2} .$ Let the angle between $\vec{a} a n d \vec{b}$ be acute, $| \vec{a} | = 1, a n d | \vec{a} . \vec{b} | = | \vec{a} * \vec{b} | .$ If $\vec{c} = 2 \sqrt[]{2} (\vec{a} * \vec{b}) - 2 \vec{b},$ then an angle between $\vec{b} a n d \vec{c} i s$

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Payal Gupta

Contributor-Level 10

Be the vectors along the diagonals of a parallelogram having are 2.

$∴ \frac{1}{2} | \vec{a} * \vec{b} | = 2 \sqrt[]{2}$

$| \vec{a} | | \vec{b} | s i n θ = 4 \sqrt[]{2}$

$\Rightarrow | \vec{b} | s i n θ = 4 \sqrt[]{2}$ ……. (i)

And

$\Rightarrow \vec{c} . \vec{b} = - 2 {| \vec{b} |}^{2} = - 128 . . . . . . (i i)$

$\Rightarrow | \vec{c} | = 16 \sqrt[]{2} . . . . . . . (i i i)$

From (ii) and (iii)

$| \vec{c} | | \vec{b} | c o s α = - 128$

$\Rightarrow c o s α = \frac{1}{\sqrt[]{2}}$

$α = \frac{3 π}{4}$

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